Important Questions Class 10 Maths Chapter 13

Important Questions Class 10 Mathematics Chapter 13 – Surface Areas and Volumes 

Mathematics can be tough only if you don’t know the right strategy and approach to study it. It is applicable in almost all domains of life and hence becomes an important subject in school as well as for competitive examinations.

The chapter surface areas and volumes is all about calculating the areas and the volumes of different shapes like cubes, cuboids, cylinders, spheres, cones, etc. The essential topics covered in this chapter include the surface area of the combination of solids, volume of the combination of solids, conversion of solid from one shape to another, and frustum of a cone.

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Students can find a lot of practice questions on the surface area and volume from our Important Questions Class 10 Mathematics Chapter 13. This will help you understand the concepts and learn the formulas with ease. You will become a smart learner and will be able to do exceedingly well in your examinations.

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Important Questions Class 10 Mathematics Chapter 13 – With Solutions

The following important questions and their solutions are included in the Class 10 Mathematics Chapter 13 important questions:

 Question 1: A cylindrical pencil sharpened at the one edge has the combination of

(A) A cone and a cylinder

(B)  frustum of a cone and a cylinder

(C) a hemisphere and a cylinder(D) two cylinders

Answer 1: (A) The cone and the cylinder

Explanation: A pencil has the combination of

The nip of the sharpened pencil is in a conical shape, and the rest of the part is cylindrical; hence the pencil is the combination of a cylinder and a cone.

So, the correct answer is option (A).

 

Question 2: A surahi has the combination of

(A) a sphere as well as a cylinder

(B) a hemisphere as well as a cylinder

(C) the two hemispheres

(D) a cylinder as well as a cone.

Answer 2: (A) a sphere as well as a cylinder

Explanation: The top part of the surahi is in a cylindrical shape and  the bottom part is in a spherical shape; hence surahi is a combination of a Sphere and a cylinder.

So the correct answer is option (A).

 

Question 3: A plumbline (Sahul) has the combination of (see Fig. )

(A) a cone as well as a cylinder

(B) a hemisphere as well as a cone

(C) frustum of the cone as well as a cylinder

(D) the sphere as well as cylinder

Answer 3: (B) a hemisphere as well as a cone

Explanation: The upper part of the plumbline is hemispherical, and the bottom part is conical. Hence, it is a combination of a hemisphere as well as a cone.

Hence, the correct answer is option (B).

 

Question 4:The shape for a glass (tumbler) (see Fig.  ) is usually of the form of

(A) The cone

(B) frustum of the cone

(C)  the cylinder

(D) the Sphere

Answer 4: (B) frustum of the cone

Explanation: The shape of the frustum of the cone is:

Hence, the shape for glass is a frustum [an inverted frustum].

Here, the correct answer is option (B).

  

Question 5: The shape for a Gilli, in the Gilli-danda game (see Fig.), has a combination of

(A) the two cylinders

(B) a cone as well as a cylinder

(C) two cones as well as a cylinder

(D) two cylinders as well as a cone

Answer 5: (C) two cones as well as a cylinder

Explanation: A Gilli has a combination of

The left and the right part for a Gilli are conical and  the central part is cylindrical.

Hence, it is a combination of a cylinder and two cones. 

So, the correct answer is option (C). 

 

Question 6:A shuttle cock used for playing badminton has the shape for the combination of

(A) a cylinder as well as a sphere

(B) a cylinder as well as a hemisphere

(C) a sphere as well as a cone

(D) frustum of the cone as well as a hemisphere

Answer 6: (D) frustum of the cone as well as a hemisphere

Explanation: The cork of the shuttle is hemispherical in shape as well as the upper part is in the shape of the frustum of a cone. Hence, it is a combination of frustum of a cone and a hemisphere.

Thus, the correct answer is Option (D).

 

Question 7: A cone intersects through a plane parallel to its base as well as then the cone that is formed at one side for the plane is removed. The new part which is left over on the other side of the plane is called.

(A) a frustum of a cone

(B) cone

(C) cylinder

(D) sphere

Answer 7:(A) a frustum of a cone

Explanation: if a cone is divided with two parts with a plane through any point on the axis parallel to the base, the upper and lower parts obtained are cones as well as a frustum, respectively.

So, the correct answer is option (A).

 

Question 8: A hollow cube with an internal edge of 22 cm is filled with spherical marbles of the diameter of 0.5 cm as well as it is assumed that 1/8 of space for the cube remains unfilled. The number of marbles which the cube can use is

(A) 142244

(B) 142344

(C) 142444

(D) 142544

Answer 8: (A) 142244

Explanation:

Diameter of  the marble = 0.5 cm

Radius = .5/2 = 5/ 20 = 1/ 4 cm 

volume of  the marble =(4πr3)/3 = 4/3 *22/7 * ¼ * ¼ * ¼ = 11/168 cm3

Edge for the cube = 22 cm

Volume (V)= 22 * 22 * 22

Space occupied with marble = total volume 1/8 part of the volume = v – v/8 = 7v/8

Number of the marbles = space occupied/ volume of marble

 =(7v * 168)/ 8*11

=  (7*22*22*22*22*168)/(8*11)

= 142244 

 

Question 9: A metallic spherical shell for internal and external diameters 4 cm and 8 cm, respectively, is melted as well as recast to form a cone of the base diameter of 8 cm. The height of  the cone is

(A) 12 cm

(B) 14 cm

(C) 15 cm

(D) 18 cm

Answer 9: (B) 14cm

Explanation: volume of  the spherical shell = volume of  the cone recast by melting

In case of spherical Shell,

Internal diameter is d1 = 4 cm

Internal radius is  r1 = 2 cm

[ as the radius = 1/2 diameter]

External diameter is d2 = 8 cm

External radius is r2 = 4 cm

Then,

When the volume of  spherical shell= 4/3 π (r2 3 – r13)

here r1 and r2 are internal and external radii respectively.

volume of  given shell = 4/3 π (43 – 23)

= 4/3 π (56)

= (224/3) π

We find

volume of  the cone = 224π /3 cm3

In the case of cone,

For Base diameter = 8 cm

Base radius is r = 4 cm

Let height of  the cone = ‘h’.

We get,

volume of  the cone = (1/3) π r2h,

here r = Base radius and h = height of cone

volume of  the given cone = (1/3) π 42h

224π /3 = 16πh /3

 16h = 224

h = 14 cm

Thus, height of  the cone is 14 cm. 

 

Question 10: A solid piece for iron in the form of a cuboid having dimensions 49 cm × 33 cm × 24 cm is molded to form the solid Sphere. The radius of  the Sphere is

(A) 21cm

(B) 23 cm

(C) 25 cm

(D) 19 cm

Answer 10: (A) 21cm

Explanation:

length of  cuboid = 49 cm

Breadth for cuboid = 33 cm

height of  cuboid = 24 cm

It is given by,

Volume = l * b * h

= 49*33*24 = 38808

We now know that volume of  the sphere  (4πr3)/3

For the volume of the sphere = 38808 (given)

r3 = (38808 *7*3)/(4*22) = 882*7*3)/2 = 9261

r  = 3√(3*3*3*7*7*7) =3*7 = 21 

Thus, the radius of the Sphere = 21 cm 

 

Question 11: A mason constructs a wall for the dimensions 270 cm × 300 cm × 350 cm by the bricks, each having a size of 22.5 cm × 11.25 cm × 8.75 cm as well as it is assumed that 1/8 space is covered with the mortar. The number for bricks used to construct the wall is

(A) 11100

(B) 11200

(C) 11000

(D) 11300

Answer 11: (B) 11200 bricks

Explanation:

length of  the wall = 270 cm

Breadth for the wall = 300 cm

Height of the wall = 350 cm

We have,

Volume = l * b * h

=270 * 300 * 350 = 28350000cm3  

length of  the brick = 22.5 cm

Breadth for the brick = 11.25 cm

height of  the brick = 8.75 cm

We get,

Volume = l * b * h  

=22.5*11.25*8.75 = 2214.84375cm3  

 1/8 Space is covered with mortar (given)

The remaining space = volume of wall /8

2835000/8 = 3543750cm3    

The surface constructed is = 28350000 – 3543750 = 24806250cm3 

Number of the bricks used = surface contracted/volume of brick

=24806250/2214.84375 = 11200  bricks

 

Question 12: Twelve solid spheres for the same size are made with melting a solid metallic cylinder with a base diameter of 2 cm and a height of 16 cm. The Diameter of  each Sphere is

(A) 4 cm

(B) 3 cm

(C) 2 cm

(D) 6 cm

Answer 12: (C) 2 cm

Explanation:

Diameter of  the metallic cylinder = 2 cm

Radius is = 2/2 = 1 cm  

Height is = 16cm

We have,

Volume =πr2h

= π*1*1*16 = 16π

We are aware that the twelve solid sphere are made with melting of solid metallic cylinder

volume of  the sphere = (4πr3)/3  

So, the volume of  the 12 spheres = 16π  

= 12*(4πr3)/3 = 16π  

= 16πr3 = 16π  

r3 = 1

r = 1

Radius is = 1 cm

Diameter is 2*1 = 2 cm    

 

Question 13: The radii for the top as well as the bottom of a bucket for the slant height of 45 cm are 28 cm and 7 cm, respectively. The curved surface area of the bucket is

(A) 4950 cm2

(B) 4951 cm2

(C) 4952 cm2

(D) 4953 cm2

Answer 13: (A) 4950 cm2

Explanation:

Slant height of  the bucket = 45 cm

Top radius is = r1  = 28cm 

Bottom radius is = r2  = 7cm  +

Curved surface area of the bucket is = πl(r1+r2)  

=22/7 * 45 * (28+7)

= 22/7 *45 *35

=4950cm2 

 

Question 14: A medicine-capsule is having the shape of a cylinder with the diameter of 0.5 cm by two hemispheres stuck to each for its ends. The length of  the entire capsule is 2 cm. The capacity for the capsule is

(A) 0.36 cm3

(B) 0.35 cm3

(C) 0.34 cm3

(D) 0.33 cm3

Answer 14: (A) 0.36 cm3

Explanation:

Diameter of  the hemisphere = 0.5 cm

Radius = 0.5/2 = 5/20 = ¼ = 0.25 cm

Volume of hemisphere = (2πr3)/3

=(2/3)*(22/7)*(1/4)*(1/4)*(1/4) =11/336

 Volume of two hemispheres = (2*11)/336 = 11/168

Same as, the radius of  the cylinder = 0.25

Height is = 2- 0.25 -0.25

= 2- 0.5

=1.5 cm

We get, volume = πr2h 

=  (22/7)*(1/4)*(1/4)*(15/10)  = 33/112

The total volume of  the capsule = volume of  two hemispheres + volume of  the cylinder

= (11/168)+(33/112)

=0.065 + 0.294

= 0.359cm3

=0.36cm3 (approximate) 

 

Question 15: If two solid hemispheres for the same base radius r are joined together along with their bases, the curved surface area of this new solid is

(A)   4πr2

(B)   6πr2

(C)   3πr2

(D)  8πr2

Answer 15: (A)   4πr2

Explanation: The radius of  the hemisphere = r

We know curved surface area = 2πr2  

The curved surface area of two solid hemisphere

= 2 * 2πr2

= 4πr2 

 

Question 16: A right circular cylinder having the radius r cm as well as height h cm (h > 2r) just encloses the Sphere of diameter

(A) r cm

(B) 2r cm

(C) h cm

(D) Two h cm

Answer 16: (B) 2r cm

Explanation: Since the Sphere is just enclosed in a cylinder,

The Diameter of the Sphere will be equal to the diameter of the cylinder.

Diameter of  the base of cylinder = 2(radius of base) = 2r

Thus, the Diameter of the Sphere = 2r cm

Therefore, the correct answer is option (B).  

 

Question 17: During the conversion of the solid from one shape to another, the volume of  the new shape will be

(A) increase

(B) decrease

(C) remains unaltered

(D) be doubled

Answer 17: (C) remains unaltered.

Explanation: When a solid is converted from one shape to another, the volume of  the new shape remains the same.

Thus, the correct answer is option (C). 

 

Question 18: The diameters for the two circular ends of the bucket are 44 cm and 24 cm. The height of  the bucket is 35 cm. The capacity for the bucket is

(A) 32.7 litres

(B) 33.7 litres

(C) 34.7 litres

(D) 31.7 litres

Answer 18: (A) 32.7 litres

Explanation:

volume of  the frustum of cone =(πh/3) *(r12+r22+r1r2)  

Diameter of  the first end = 44 cm 

Radius is r1 = 44/2 = 22cm 

Diameter of  the second end = 24 cm

Radius is r2 = 24/2 = 12cm   

For Height (h) = 35cm

Volume is = (πh/3) *(r12+r22+r1r2)  

= 22/7 * 35/3 * (222+122+22*12)

 =110/3(484+144+264)

 =(110/3) * (892)

 =32706.67cm3 

We now know that 1 liter = 1000 cm3

Therefore, 32706.67cm3  = 32.7l

 

Question 19: In the right circular cone, the cross-section made with a plane parallel to the base is as follows:

(A) circle

(B) frustum of a cone

(C) sphere

(D) hemisphere

Answer 19: (B) frustum of a cone

Explanation:  As per the question, when a right circular cone is cut with a plane parallel to the base, the figure formed is

Where BECD is not a circle, not a sphere, not a hemisphere but is instead a frustum of a cone.

Therefore, in a right circular cone, the cross-section made with a plane parallel to the base is a frustum of a cone.

 

Question 20: Volumes for two spheres are in the ratio 64: 27. The ratio for their surface areas is

(A) 3: 4

(B) 4 : 3

(C) 9: 16

(D) 16: 9

Answer 20: (D) 16 : 9

Explanation: Assume two-sphere having radius r1 and r2

As per the question, 

volume of the first sphere / volume of the second sphere = 64/27

= (4/3 *πr13)/(4/3 *πr23) = 64/27

(r1/r2)3 = 64/27

r1/r2  = 3√(64/27) =4/3

 Ratio for their surface area is = (4 *πr12)/(4 *πr22)  = r12/r22  = (r1/r2)2 =  (4 /3)2 = 16/9

 So, the required ratio is 16: 9

 

Question 21. Two identical solid hemispheres for equal base radius r cm are stuck together along their bases. The total surface area of the combination is 6πr2.

Answer 21: False

Explanation: 

If two hemispheres are joined together along with their bases, a sphere for the same base radius is formed.

So, Curved Surface area of a sphere = 4πr2.

 

Question 22. A solid cylinder for the radius r and height h is placed over other cylinders of the same height and radius. The total surface area of the shape so formed is 4πrh + 4πr2.

Answer 22: False

Explanation:

As per the question,

If one cylinder is placed over another, the base for the first cylinder and the top for the other cylinder will not be covered in the total surface area.

We get,

Total surface area of the cylinder = 2πrh + 2πr2h, here r = base radius and h = height

Total surface area of the shape formed = 2(Total surface of single cylinder) – 2(Area of base of cylinder)

= 2(2πrh + 2πr2) – 2(πr2)

= 4πrh + 2πr2

Question 23. A solid ball is perfectly fitted inside the cubical box for the side a. The volume of  the ball is 4/3πa3.

Answer 23:False

Explanation:

Assume the radius of  the sphere = r

If a solid ball is exactly fitted inside the cubical box for the side a,

We find,

Diameter of  the ball = Edge length of the cube

2r = a

For Radius, r = a/2

We get,

volume of  the Sphere = 4/3πr3

volume of  the ball 

= 4/3π(a/2)3 

= 4/3π(a3/8) 

= 1/6πa3

 

Question 24. Two identical cubes, each having a volume of 64 cm3, are joined together end to end. What is the surface area of the resulting cuboid?

Answer 24:

Assume the side of one cube = a

Surfaces area of the resulting cuboid 

= 2(Total surface area of a cube) – 2(area of a single surface)

We get,

Total surface area of the cube = 6a2 , here a = side of cube

 Surfaces area of the resulting cuboid = 2(6a2) – 2(a2) = 10a2

And,

As per the question,

volume of  the cube = 64 cm3

For the Volume of cube, we have = a3

64 = a3

a = 4 cm

Hence,

Surface area of the resulting cuboid 

= 10a2

= 10(4)2 

= 160 cm2

 

Question 25. From the solid cube for a side of 7 cm, a conical cavity for the height of 7 cm as well as a radius of 3 cm is hollowed out. Find out the volume of the remaining solid.

Answer 25:

From the above figure, we find,

volume of  the remaining solid 

= volume of  the cube – volume of  the cone

In case of Cube

Side is a = 7 cm

We get,

volume of  the cube = a3, here a = side of cube

volume of  the cube = (7)3 = 343 cm3

In case of cone

Radius is, r = 3 cm

Height is, h = 7 cm

volume of  the cone 

= 1/3 πr2h 

= 1/3 π(3)27 

= 3 × (22/7) × 7 

= 66 cm3

volume of  remaining solid 

= volume of  the cube – volume of  the cone

= 343 – 66

= 277 cm3

 

Question 26. 2 cubes each for the volume of 64 cm3 are joined end to end. Find out the surface area of the resulting cuboid.

Answer 26:

The diagram is given as follows:

Given that,

The Volume (V) for each cube is = 64 cm3

It implies that a3 = 64 cm3

Hence, a = 4 cm

Then, the side for the cube = a = 4 cm

And the length and breadth for the resulting cuboid will be of 4 cm each whereas its height will be of 8 cm.

Hence, the surface area of the cuboid = 2(lb+bh+lh)

= 2(8×4+4×4+4×8) cm2

= 2(32+16+32) cm2

= (2×80) cm2 = 160 cm2

 

Question 27. A vessel is of the form for a hollow hemisphere mounted with a hollow cylinder. The Diameter of  the hemisphere is 14 cm as well as the total height of the vessel is 13 cm. Find out the inner surface area of the vessel.

Answer 27:

The diagram given below is:

Then, the given parameters are as follows:

The Diameter of  the hemisphere = D = 14 cm

The radius of  the hemisphere = r = 7 cm

And the height of  the cylinder = h = (13-7) = 6 cm

Also, the radius of the hollow hemisphere = 7 cm

Then, the inner surface area of the vessel = CSA for the cylindrical part + CSA for the hemispherical part

(2πrh+2πr2) cm2 = 2πr(h+r) cm2

2×(22/7)×7(6+7) cm2 = 572 cm2

 

Question 28. A cubical block of the side 7 cm is surmounted with a hemisphere. What is the greatest diameter the hemisphere could have? Find out the surface area of the solid.

Answer 28:

It is given that each side for the cube is 7 cm. Hence, the radius will be 7/2 cm.

We get,

The total surface area of the solid (TSA) 

= surface area of the cubical block + CSA for the hemisphere – area of the base of the hemisphere

So, TSA of the solid = 6×(side)2+2πr2-πr2

= 6×(side)2+πr2

= 6×(7)2+(22/7)×(7/2)×(7/2)

= (6×49)+(77/2)

= 294+38.5 = 332.5 cm2

Hence, the surface area of the solid is 332.5 cm2

 

Question 29. A hemispherical depression is cut off from one face for a cubical wooden block so that the diameter l for the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Answer 29:

The diagram is given below:

Then, the Diameter of  the hemisphere = Edge of the cube = l

Hence, the radius of  the hemisphere = l/2

Thus, The total surface area of the solid 

= surface area of the cube + CSA for the hemisphere – area of the base of the hemisphere

TSA for the remaining solid 

= 6 (edge)2+2πr2-πr2

= 6l2 + πr2

= 6l2+π(l/2)2

= 6l2+πl2/4

= l2/4(24+π) sq. units

 

Question 30. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each for its ends. The length of the entire capsule is 14 mm, and the Diameter of the capsule is 5 mm. Find out its surface area.

Answer 30:

Two hemispheres, as well as one cylinder, are shown in the figure given below.

Where the Diameter of  the capsule = 5 mm

So, the Radius = 5/2 = 2.5 mm

Then, the length of  the capsule = is 14 mm

Hence, the length of  the cylinder = 14-(2.5+2.5) = 9 mm

Therefore, The surface area of the hemisphere = 2πr2 = 2×(22/7)×2.5×2.5

= 275/7 mm2

Then, the surface area of the cylinder = 2πrh

= 2×(22/7)×2.5×9

(22/7)×45 = 990/7 mm2

So, the required surface area of the medicine capsule will be

= 2×surface area of the hemisphere + surface area of the cylinder

= (2×275/7) × 990/7

= (550/7) + (990/7) = 1540/7 = 220 mm2

 

Question 31. the tent is in shape for the cylinder surmounted by a conical top. When the height and Diameter of  the cylindrical part are 2.1 m and 4 m, respectively, as well as the slant height of  the top, is 2.8 m, find out the area of the canvas used for making the tent. Also, find out the cost of the canvas for the tent at the rate of Rs 500 per m2. (Note that the base for the tent will not be covered with canvas.)

Answer 31:

It is known that a tent has a combination of a cylinder and a cone.

We find,

Diameter = 4 m

Slant height of  the cone (l) = 2.8 m

radius of  the cone (r) = radius of  the cylinder = 4/2 = 2 m

height of  the cylinder (h) = 2.1 m

Hence, the required surface area of the tent = surface area of the cone + surface area of the cylinder

= πrl+2πrh

= πr(l+2h)

= (22/7)×2(2.8+2×2.1)

= (44/7)(2.8+4.2)

= (44/7)×7 = 44 m2

Hence, The cost of the canvas of the tent at the rate for the ₹500 per m2 will be

= Surface area × cost per m2

 = ₹22000

Therefore, Rs. 22000 will be the total cost of the canvas.

 

Question 32. A solid is in shape for the cone standing on the hemisphere with both their radii being equal to the 1 cm, and the height of  the cone is equal to its radius. Find out the volume of the solid in terms of π.

Answer 32:

We have,

we know r = 1 cm and h = 1 cm.

In the diagram given below,

Then, volume of  the solid = volume of  the conical part + volume of  the hemispherical part

We now know the volume of  the cone = ⅓ πr2h

Also,

The volume of  the hemisphere = ⅔πr3

Hence, the volume of  the solid will be 

=π/3 *(1)2 [1 + 2(1)]cm3

= π/3 *(1) *3cm3

 = π cm3

 

Question 33. A metallic sphere for the radius of 4.2 cm is melted as well as recast into the shape of a cylinder for the radius of 6 cm. Find out the height of the cylinder.

Answer 33:

We have a radius of  the Sphere (R) = 4.2 cm

And, radius of  the cylinder (r) = 6 cm

Then, let the height of  the cylinder = h

It is given that the Sphere is converted into a cylinder.

Therefore, volume of  the Sphere = volume of  the cylinder

Hence, (4/3)×π×R3 = π×r2×h.

h = 2.74 cm

 

Question 34. The spherical glass vessel has a cylindrical neck of 8 cm long and 2 cm in diameter; the Diameter of  the spherical part is 8.5 cm. By measuring the amount for water it holds, a child finds the volume to be 345 cm3. Check if she is correct, considering the above as the inside measurements, and π = 3.14.

Answer 34:

Given that,

In case of the cylinder part, Height (h) = 8 cm and Radius (R) = (2/2) cm = 1 cm

In case of the spherical part, Radius (r) = (8.5/2) = 4.25 cm

So, volume of  this vessel = volume of  the cylinder + volume of  the sphere,

= π×(1)2×8+(4/3)π(4.25)3

= 346.51 cm3

Question 35: The canal is 300 cm wide as well as 120 cm deep. The water in the canal is flowing with the speed of 20 km/h. How much area would it irrigate in 20 minutes when 8 cm for the standing water is desired?

Answer 35:

The volume of  the water flows in the canal in one hour = width for the canal × depth for the canal × speed for the canal water 

= 3 × 1.2 × 20 × 1000 m3

= 72000 m3

In 20 minutes, the volume of  the water 

= (72000 × 20)/ 60 = 24000 m3

Area irrigated in 20 minutes, when 8 cm, that is 0.08 m standing water is needed

=24000/0.08

= 300000 m2

= 30 hectares

 

Question 36: Two cones have their heights in the ratio of 1 : 3 as well as radii in the ratio of 3: 1. What is the ratio for their volumes?

Answer 36:

Given that,

Ratio for the heights of two cones = 1 : 3

Ratio for the radii = 3 : 1

Let h and 3h be the height of  the two cones.

And, 3r and r be the corresponding radii for the cones.

Hence,, r1 = 3r, h1 = h, r2 = r, h2 = 3h.

Ratio for the volumes = [(1/3)πr12h1]/ [(1/3)πr22h2]

= [(3r)2 h]/[r2 (3h)]

= (9r2h)/(3r2h)

= 3/1

Thus, the ratio of volumes = 3 : 1

 

Question 37: A cubical ice-cream brick for the edge of 22 cm is to be distributed between some children by filling ice-cream cones of a radius of 2 cm as well as a height of 7 cm up to its brim. How many children would get the ice cream cones? 

Answer 37:

Let n be the number for the ice-cream cones.

volume of  the cubical ice-cream brick = 22 cm × 22 cm × 22 cm

radius of  the cone = r = 2 cm

height of  the cone = h = 7 cm

volume of  the cone = (1/3)πr2h = (1/3) × (22/7) × 2 × 2 × 7

So, 

n × volume of  the one cone = volume of  the cubical ice-cream brick

n × (1/3) × (22/7) × 2 × 2 × 7 = 22 × 22 × 22

n × (1/3) × 4 = 22 × 22

n = (22 × 22 × 3)/4

n = 363

Hence, 363 children will get the ice cream cones.

 

Question 38: Three cubes for the metal those edges are in the ratio of 3:4:5 are melted as well as converted into a single cube that has a diagonal of 12√3 cm. Find out the edges of the three cubes.

Answer 38:

Let the edges for the three cubes (in cm) by 3x, 4x, and 5x, respectively.

volume of  the the cubes after melting is = (3x)3 + (4x)3 + (5x)3 = 216×3 cm3

Let a be the side for the new cube so formed after melting. 

Hence,, a3 = 216×3

Thus,, a = 6x

Given that, the diagonal for the single cube = 12√3 cm.

i.e. √(a2 + a2 + a2) = 12√3

a√3 = 12√3

Hence, a = 12

So, 12 = 6x

x = 2

Then, 3x = 3 × 2 = 6

4x = 4 × 2 = 8

5x = 5 × 2 = 10

here, the edges of the three cubes are 6 cm, 8 cm, and 10 cm, respectively.

 

Question 39: Find the number for the solid spheres, each of diameter 6 cm, so that they can be made by melting a solid metal cylinder for the height of 45 cm and a diameter of 4 cm.

Answer 39:

Given that,

Diameter of  the solid sphere = 6 cm

Diameter of  the cylinder = 4 cm

height of  the cylinder = h = 45 cm

radius of  the sphere = r1= 6/2 = 3 cm

radius of  the cylinder = r2 = 4/2 = 2 cm

Assume n be the number of spheres.

n × volume of  the one sphere = volume of  the cylinder

n × (4/3)πr13 = πr22h

n × (4/3) × (22/7) × 3 × 3 × 3 = (22/7) × 2 × 2 × 45

n × 9 = 45

n = 45/9

n = 5

 

Question 40. A solid cone of radius r and height h is placed over a solid cylinder having the same base radius and height as that for the cone. The total surface area of the combined solid is πr[√(r2 + h2 +3r + 2h].

Answer 40:

False

Explanation:

When a solid cone is placed over a solid cylinder for the same base radius, the base for the cone and top for the cylinder will not be covered in total surface area.

As the height of  the cone and cylinder is the same,

We find,

Total surface area of the cone = πrl + πr2, here r = base radius and l = slant height

Total surface area of the shape formed = Total surface area of the cone + Total Surface area of the cylinder – 2(area of the base)

Total surface area of the cylinder = 2πrh + 2πr2 h, here r = base radius and h = height

= πr(r + l) + (2πrh + 2πr2 ) – 2(πr2 )

= πr2  + πrl + 2πrh + 2πr2  – 2πr2 

= πr(r + l + h)

= πr(r + √(r2 + h2)+ 2h)

 

Question 41. Three metallic solid cubes whose edges are 3 cm, 4 cm, and 5 cm are melted into and formed into a single cube. Find out the edge of the cube so formed.

Answer 41:

We get,

volume of  the cube = a3, here a = side of cube

As per the question,

Side for the first cube, a1= 3 cm

Side for the second cube, a2 = 4 cm

Side for the third cube, a3= 5 cm

Assume that the side for the cube recast from melting these cubes = a

We know that the total volume of  the 3 cubes will be the same as the volume of  the newly formed cube,

volume of  the new cube = (volume of  the 1st + 2nd + 3rd cube)

⇒ a3 = a13 + a23 + a33

⇒ a3 = (3)3 + (4)3 + (5)3

⇒ a3 = 27 + 64 + 125 = 216

⇒ a = 6 cm

So, side for the cube so formed is 6 cm.

 

Question 42. How many shots, each having a diameter of 3 cm, can be made from a cuboidal lead solid for the dimensions 9cm × 11cm × 12cm?

Answer 42:

volume of  cuboid = lbh, here, l = length, b = breadth and h = height

For Cuboidal lead:

Length, l = 9 cm

Breadth, b = 11 cm

Height, h = 12 cm

volume of  lead = 9(11)(12) = 1188 cm3

volume of  sphere = 4/3πr3, where r = radius of sphere

For Spherical shots,

Diameter = 3 cm

Radius, r = 1.5 cm

volume of  shot = 4/3 × 22/7 × (1.5)3 = 99/7 cm3

No. of shots can be made = volume of lead / volume of one shot = 1188/(99/7) =1188*7/99 = 84

 

Therefore, the number of bullets that can be made from lead = 84.

 

Question 43. A bucket is in the form for the frustum of a cone and holds 28.490 litres of water. The radii for the top and bottom are 28 cm and 21 cm, respectively. Find out the height of the bucket.

Answer 43:

As per the question,

The bucket is in the form of frustum for the cone.

We get,

Volume of frustum for the cone = 1/3 πh(r12 + r22 + r1r2), here, h = height, r1and r2 are the radii(r1 > r2)

For bucket,

volume of  the bucket = 28.490 L

1 L = 1000 cm3

volume of  the bucket = 28490 cm3

radius of  the top, r1 = 28 cm

radius of  the bottom, r2 = 21 cm

Let the height = h.

Putting these values in the equation to find the volume of the bucket,

We get,

volume of  the bucket = 1/3 πh[282 + 212 + 28(21)]

28490 = 1/3 × 22/7 × h (784 + 441 + 588) = 22/7 × h × 1813

⇒ h = (28490 × 21) / (22 × 1813)

⇒ h = 15

 

Question 44. A solid metallic hemisphere of radius 8 cm is melted and recast into a right circular cone for the base radius of 6 cm. Find out the height of the cone.

Answer 44:

For hemisphere,

Radius, r = 8 cm

We get,

volume of  the hemisphere = 2/3 πr3. Here, r = radius of the hemisphere

Thus, we get,

volume of  the given hemisphere = 2/3 × π × 83 = (1024/3) π cm3

Now,

For the cone that is recast from a hemisphere,

Base radius, r = 6 cm

We get,

volume of  the cone = 1/3 πr2h. Here, r is the base radius, and h is the height of the cone.

Thus, we get,

volume of  the cone = 1/3 π(6)2h = 12πh

From the question, we get,

The volume remains the same if a body is reformed to another body

volume of  the cylinder = volume of  the cone

12πh = 1024π /3

h = 28.44 cm

 

Question 45. A rectangular water tank of base 11 m × 6 m has water up to a height of 5 m. When the water in the tank is transferred to a cylindrical tank for the radius of 3.5 m, find out the height of the water level in the tank.

Answer 45:

volume of  the water in the tank = volume of  the cuboidal tank up to a height of 5 m

As per the question,

Consider cuboidal tank

Length, l = 11 m

Breadth, b = 6 m

Height, h = 5m

We know that the equation to find the volume of   the tank,

volume of  the tank = lbh. Here, l, b, and h are the length, breadth, and height of the tank, respectively

volume of  the water = 11(6)(5) = 330 m3

We get,

Base radius of  the cylindrical tank, r = 3.5 m

Assume the height till which the cylindrical tank is filled = h m

 Using the formula,

volume of  the cylinder = πr2h, here r is base radius, and h is the height of cylinder

volume of  the water in cylindrical tank = π(3.5)2h

330 m3 = 22/7 × 3.5 ×3.5 × h

330 m3 = h × 38.5

h = 8.57 m

So, the height till which the cylindrical tank is filled = 8.57 m

 

Question 46. A factory manufactures 120000 pencils daily. The pencils are cylindrical in shape, each with a length of 25 cm and a circumference for the base of 1.5 cm. Find out the cost of coloring the curved surfaces of the pencils manufactured in one day at Rs 0.05 per dm2.

Answer 46:

The shape for the pencil is a cylinder.

Let the radius of  the base = r cm

Circumference for the base = 1.5 cm

Circumference for the circle is 2πr = 1.5 cm

r = 1.5/2π cm

As per the question,

Height, h = 25 cm

We get,

Curved surface area of the cylinder = 2πrh

Curved surface area of the pencil = 2π (1.5/2π) 25 = 37.5 cm2

1 cm = 0.1 dm

1 cm2 = 0.01 dm2

37.5 cm2 = 0.375 dm2

Cost of coloring 1 dm2 = Rs. 0.05

Cost of coloring 0.375 dm2 (i.e. 1 pencil) = Rs. 0.01875

Cost of coloring 120000 pencils = 120000 ×0.01875 = Rs. 2250

 

Question 47. Five hundred persons are taking a dip into a cuboidal pond for which is 80 m long and 50 m broad. What is the rise of the water level in the pond when the average displacement for the water by a person is 0.04m3?

Answer 47:

As per the question,

Average displacement for the person = 0.04 m3

Average displacement for the 500 persons = 500 × 0.04 = 20 m3

So, the volume of  the water raised in the pond = 20 m3

We have,

length of  the pond, l = 80 m

Breadth for the pond, b = 50 m

Height = h

volume of  the water raised in pond = 80(50)(h)

20 m3 = 4000h

h = 0.005 m = 0.5 cm

So, raise the height of  the water = 0.5 cm.

Question 48. The toy is in the form of the cone for the radius of 3.5 cm mounted on a hemisphere for the same radius. The total height of  the toy is 15.5 cm. Find out the total surface area of the toy.

Answer 48:

The diagram is as:

Given that the radius of  the cone and the hemisphere (r) = 3.5 cm or 7/2 cm

The total height of  the toy is given as 15.5 cm.

So, the height of  the cone (h) = 15.5-3.5 = 12 cm

Slant height of the cone(l) =  √(h2 + r2)

= l = √(122 + 3.52)

= l = √(122 + (7/2)2)

= l = √(144 + (49/4))

= l = √(576 +49)/4)

= l = √(625)/4

= l = 25/2

 Hence, The curved surface area of the cone = πrl

(22/7)×(7/2)×(25/2) = 275/2 cm2

And the curved surface area of the hemisphere = 2πr2

2×(22/7)×(7/2)2

= 77 cm2

Total surface area of the toy = CSA for the cone + CSA for the hemisphere

= (275/2)+77 cm2

= (275+154)/2 cm2

= 429/2 cm2 = 214.5cm2

Hence, the total surface area (TSA) of the toy is 214.5cm2

 

Question 49. From a solid cylinder those height is 2.4 cm as well as diameter of 1.4 cm. A conical cavity for the

same height as well as same diameter is hollowed out. Find out the total surface area of the

remaining solid to the nearest cm2.

Answer 49:

The diagram is as follows:

From the question we have:

The Diameter of  the cylinder = Diameter of  the conical cavity = 1.4 cm

Hence, the radius of  the cylinder = radius of  the conical cavity = 1.4/2 = 0.7

And the height of  the cylinder = height of  the conical cavity = 2.4 cm

Therefore, Slant height of the conical cavity (l) =  √(h2 + r2)

= l = √(2.42 + 0.72) 

= l = √(5.76 +0.49) 

= l = √(6.25) 

= l = 2.5 cm

 TSA for the remaining solid = surface area of the conical cavity + TSA for the as the cylinder

= πrl+(2πrh+πr2)

= πr(l+2h+r)

= (22/7)× 0.7(2.5+4.8+0.7)

= 2.2×8 = 17.6 cm2

Hence, the total surface area of the remaining solid is 17.6 cm2

 

Question 50. A pen stand made by the wood is in the shape for the cuboid with four conical depressions to hold pens. The dimensions for the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each for the depressions is 0.5 cm, and the depth is 1.4 cm. Find out the volume of wood in the entire stand (see Fig.).

Answer 50:

volume of  the cuboid = length x width x height

We get cuboid’s dimensions as 15 cmx10 cmx3.5 cm

So, the volume of  the cuboid = 15x10x3.5 = 525 cm3

We have, depressions are like cones and we know,

volume of  the cone = (⅓)πr2h

Given that, radius (r) = 0.5 cm and depth (h) = 1.4 cm

So,  volume of  the 4 cones = 4x(⅓)πr2h

= 1.46 cm2

volume of  the wood = volume of  the cuboid – 4 x volume of cone

= 525-1.46 = 523.54 cm2

 

Question 51. A vessel is in the form for an inverted cone. that’s height is 8 cm, and the radius of  the top, which is open, is 5 cm. It is fill up with water up to the brim. If lead shots, each for the which is the sphere of radius 0.5 cm are dropped into the vessel, one-fourth for the water flows out. Find  out the number for the lead shots dropped in the vessel.

Answer 51:

In the case of cone,

Radius = 5 cm,

Height = 8 cm

Also,

radius of  the sphere = 0.5 cm

The diagram will be,

It is given that,

volume of  the cone = volume of  the water in the cone

= ⅓πr2h = (200/3)π cm3

Total volume of  the water overflown= (¼)×(200/3) π =(50/3)π

volume of  the lead shot

= (4/3)πr3

= (1/6) π

The number for the lead shots = Total volume of  the water overflown/volume of  the Lead shot

= (50/3)π/(⅙)π

= (50/3)×6 = 100 leadshots

 

Question 52. A solid consisting for the right circular cone of height of 120 cm and radius of 60 cm standing on a hemisphere for the radius of 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find out the volume of  the water left in the cylinder when the radius of  the cylinder is 60 cm, and its height is 180 cm.

Answer 52:

The volume of  the water left will be = volume of  the cylinder – volume of  the solid.

Given that,

radius of  the cone = 60 cm,

height of  the cone = 120 cm

radius of  the cylinder = 60 cm

height of  the cylinder = 180 cm

radius of  the hemisphere = 60 cm

Total volume of  the solid = Volume of Cone + Volume of hemisphere

volume of  the cone = 1/3πr2h = 1/3 × π×602×120cm3 = 144×103π cm3

volume of  the hemisphere = (⅔)×π×603 cm3 = 144×103π cm3

Hence, total volume of  the solid =  144×103π cm3 + 144×103π cm3 = 288 ×103π cm3

volume of  the cylinder = π×602×180 = 648000 = 648×103 π cm3

volume of  the water left will be = Volume of cylinder – Volume of solid

= (648-288) × 103×π = 1.131 m3

 

Question 53. Metallic spheres for the radii 6 cm, 8 cm, and 10 cm, respectively, are melted to form a single solid sphere. Find out the radius of the resulting sphere.

Answer 53:

For Sphere 1:

Radius (r1) = 6 cm

Volume (V1) = (4/3)×π×r13

For Sphere 2:

Radius (r2) = 8 cm

Volume (V2) = (4/3)×π×r23

For Sphere 3:

Radius (r3) = 10 cm

Volume (V3) = (4/3)× π× r33

And, let the radius of  the resulting Sphere be “r.”

volume of  the sphere = (V1)+(V2)+(V3)

(4/3)×π×r3 = (4/3)×π×r13+(4/3)×π×r23 +(4/3)×π×r33

r3 = 63+83+103

r3 = 1728

r = 12 cm

 

Question 54. A well for the diameter of 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in shape for the circular ring of width 4 m to form an embankment. Find out the height of the embankment.

Answer 54:

The shape for the well will be cylindrical as given below.

Given that, Depth (h1) of well = 14 m

Diameter of the circular end of the well =3 m

Hence, Radius (r1) = 3/2 m

Width for the embankment = 4 m

From the given figure, it can be said that the embankment will be the cylinder having the outer radius (r2) as 4+(3/2) = 11/2 m as well as an inner radius (r1) as 3/2m

 let the height of  the embankment be h2

volume of  the soil dug from the well = volume of  the earth used to form the embankment

π×r12×(h1) = π×(r22-r12) × h2

On Solving this, we get,

The height of  the embankment (h2) as 1.125 m.

 

Question 55. The cylindrical bucket, 32 cm high and for the radius of the base of 18 cm, is filled with sand. This bucket is emptied on the ground, and a conical heap for the sand is formed. When the height of the conical heap is 24 cm, find out the radius and slant height of the heap.

Answer 55:

The diagram is given below as-

Given,

Height (h1) for the cylindrical part of the bucket = 32 cm

Radius (r1) for the circular end of the bucket = 18 cm

height of  the conical heap ((h2) = 24 cm

Let “r2” be the radius of  the circular end of the conical heap.

We have seen volume of  the sand in the cylindrical bucket will be equal to the volume of  the sand in the conical heap.

Therefore, volume of  the sand in the cylindrical bucket = volume of  the sand in conical heap

π×r12×h1 = (⅓)×π×r22×h2

π×182×32 = (⅓)×π ×r22×24

And r2= 36 cm

here,

height of Slant  (l) = √(362+242) = 12√13 cm.

 

Question 56. Water in the canal, 6 m wide and 1.5 m deep, is flowing at a speed of 10 km/h. How much area would it irrigate in 30 minutes when 8 cm of standing water is needed?

Answer 56:

we know that the canal is the shape for the cuboid with dimensions as:

Breadth (b) = 6 m and Height (h) = 1.5 m

We have 

The speed for the canal = 10 km/hr

length of  the canal covered in 1 hour = 10 km

length of  the canal covered in 60 minutes = 10 km

length of  the canal covered in 1 min = (1/60)x10 km

length of  the canal covered in 30 min (l) = (30/60)x10 = 5 km = 5000 m

We see that the canal is cuboidal in shape. So,

volume of  the canal = lxbxh

= 5000x6x1.5 m3

= 45000 m3

volume of  the water in canal = volume of  the area irrigated

= Area irrigated x Height

Hence, Area irrigated = 56.25 hectares

Thus, Volume of canal = lxbxh

45000 = Area irrigated x 8 cm

45000 = Area irrigated x (8/100)m

And, Area irrigated = 562500 m2 = 56.25 hectares.

 

 

Question 57. A farmer connects a pipe with an internal diameter of 20 cm from a canal into a cylindrical tank in her field, which is of 10 m in diameter and 2 m deep. When water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Answer 57:

Consider the diagram-

Radiums r1 of the circular end of the pipe = 20/200 = 0.1m

Area of the cross section = πr12 = π(0.1)2=0.01πm2

Speed of water = 3km/h = 3000/60 =50 meter/min

Volume of water that flows in 1 mint from the pipe = 50* 0.01π = 0.5πm3

volume of  the water that flows in t minutes from pipe = t×0.5π m3

Radius (r2) for the circular end of cylindrical tank =10/2 = 5 m

Depth (h2) for the cylindrical tank = 2 m

Assume the tank be filled completely in t minutes.

volume of  the water-filled in the tank in t minutes is equal to the volume of  the water flowing in t minutes from the pipe.

volume of  the water that flows in t minutes from the pipe = volume of  the water in the tank

t×0.5π = π×r22×h2

Or, t = 100 minutes

 

Question 58. The cistern, internally measuring 150 cm × 120 cm × 100 cm, contains 129600 cm3 of water in it. Porous bricks are placed into the water until the cistern is full to the brim. Each brick absorbs one-seventeenth for its own volume of water. How many bricks can be put inside without overflowing the water, each being 22.5 cm × 7.5 cm × 6.5 cm?

Answer 58:

Given that the dimension for the cistern = 150 × 120 × 110

Thus, volume = 1980000 cm3

Volume to be filled in the cistern = 1980000 – 129600

= 1850400 cm3

Let the number for the bricks placed to be “n.”

Thus, the volume of n bricks will be = n×22.5×7.5×6.5

Since each brick absorbs one-seventeenth for the volume, the volume would be

= n/(17)×(22.5×7.5×6.5)

From the given condition

The volume of  the n bricks has to be equal to the volume absorbed by n bricks + the volume to be filled in the cistern

And,  n×22.5×7.5×6.5 = 1850400+n/(17)×(22.5×7.5×6.5)

On Solving, we observe

that n = 1792.41

 

Question 59. In one fortnight of the given month, there was a rainfall for the 10 cm in a river valley. When the area of the valley is 7280 km2, the total rainfall was approximately equivalent to the addition to the normal water for the three rivers, each 1072 km long, 75 m wide, and 3 m deep.

Answer 59:

From the given condition,

Total volume of  the 3 rivers = 3×[(Surface area of the river)×Depth]

Given that,

Surface area of the river = [1072×(75/1000)] km

And,

Depth = (3/1000) km

volume of  the 3 rivers = 3×[1072×(75/1000)]×(3/1000)

= 0.7236 km3

volume of  the rainfall = total surface area × total height of  the rain

= 7280*10/(100*1000) km3 

= 0.7280 km3

here, the total rainfall was approximately equivalent to the sum of the normal water for the three rivers, 

The volume of  the rainfall has to be equal to the volume of 3 rivers.

If 0.7280 km3 = 0.7236 km3

Hence, the question statement was true.

 

Question 60. An oil funnel made with a tin sheet consists of the 10 cm long cylindrical portion attached to a frustum for the cone. When the total height is 22 cm, the Diameter of  the cylindrical portion is 8 cm, and the Diameter of  the top of the funnel is 18 cm. Find out the area of the tin sheet required to make the funnel (see Fig.).

Answer 60:

Given that,

Diameter of  the upper circular end of frustum part = 18 cm

Thus, radius (r1) = 9 cm

The radius of the lower circular end for the frustum (r2) will be equal to the radius of  the circular end of the cylinder

Thus, r2 = 8/2 = 4 cm

height (h1) for the frustum section = 22 – 10 = 12 cm

And,

Height (h2) for the cylindrical section = 10 cm (given)

Then, the slant height will be-

L =  √[(r1-r2)2 -h12]

And  l = 13 cm

area of the tin sheet required = CSA for the frustum part + CSA for the cylindrical part

= π(r1+r2)l+2πr2h2

On Solving, we get,

area of the tin sheet required = 782 4/7 cm2

Question 61: Rasheed gets a playing top (lattu) as his birthday present, in which, surprisingly, he found no colour on it. He wanted to colour it with his crayons. The top is shaped as like a cone surmounted with a hemisphere. The entire top of 5 cm in height, as well as the diameter of the top, is 3.5 cm. Find out the area he has to colour. (Take π = 22/7)

Answer 61:

TSA for the toy = CSA for the hemisphere + CSA for the cone

Curved surface area of the hemisphere = 1/ 2 (4πr2) = 2π r2 = 2(22/7)× (3.5/2) × (3.5/2) = 19.25 cm2

height of  the cone = height of  the top – radius of  the hemispherical part

= (5 – 3.5/2) cm = 3.25 cm

Slant height of  the cone (l)

(approx.)

Thus, CSA of cone = πrl = (22/7) × (3.5/2) × 3.7 = 20.35 cm2

So, the surface area of the top = [2(22/7)× (3.5/2) × (3.5/2) + (22/7) × (3.5/2) × 3.7] cm2

= (22/7) × (3.5/2) (3.5+3.7) cm2

= (11/2) × (3.5 + 3.7) cm2

= 39.6 cm2 (approximately)

 

Question 62: Mayank made a bird bath for his garden in the shape of a cylinder by a hemispherical depression at one end, as shown in the figure. The height of  the cylinder is 1.45 m, and its radius is 30 cm. Find out the total surface area of the bird bath. (Take π = 22/7)

Answer 62:

Assume h be the height of  the cylinder and r be the common radius of  the cylinder and hemisphere. 

The total surface area of the bird-bath = CSA for the cylinder + CSA for the hemisphere

= 2πrh + 2πr2

= 2π r(h + r)

= 2 (22/7) × 30 × (145 + 30) cm2

= 33000 cm2 = 3.3 m2

 

Question 63: 2 cubes each for the volume of 64 cm3 are joined end to end. Find out the surface area of the resulting cuboid.

Answer 63:

The diagram is as below:

Given that,

The Volume (V) for each cube is = 64 cm3

This implies that a3 = 64 cm3

The side for the cube, i.e., a = 4 cm

And the breadth and length of  the resulting cuboid will be 4 cm each, while its height will be 8 cm.

Thus, the surface area of the cuboid (TSA) = 2(lb + bh + lh)

On putting the values, we observe,

= 2(8×4 + 4×4 + 4×8) cm2

= (2 × 80) cm2

Therefore, TSA of the cuboid = 160 cm2

 

Question 64: A tent is in shape for the cylinder surmounted by a conical top. If the height and Diameter of  the cylindrical part are 2.1 m and 4 m, respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find out the cost for the canvas of the tent on the rate of Rs. 500 per m2. (Noted that the base of the tent would not be covered with the canvas.)

Answer 64:

we know that a tent is a combination for the cone and a cylinder, as given below.

From the question, we get,

For The diameter = D = 4 m

l = 2.8 m (slant height)

The radius of  the cylinder is equal to the radius of the cylinder

Hence, r = 4/2 = 2 m

And, we know the height of  the cylinder (h) is 2.1 m

Hence, the required surface area of the given tent = surface area of the cone (the top) + surface area of the cylinder (the base)

= πrl + 2πrh

= πr (l+2h)

Substituting the values as well as solving them, we observe the value as 44 m2

The cost of the canvas on the rate for Rs. 500 per m2 for the tent will be

= Surface area × cost/ m2

= 44 × 500

Therefore, Rs. 22000 will be the total cost for the canvas.

 

Question 65: The solid toy is in the form for the hemisphere surmounted by a right circular cone. The height of  the cone is 2 cm, and the diameter of the base is 4 cm. Determine the volume of  the toy. When a right circular cylinder circumscribes the toy, find out the difference between the volumes for the cylinder and the toy. (Take π = 3.14)

Answer 65:

Assume BPC be the hemisphere and ABC be the cone standing on the base for the hemisphere, as shown in the above figure.

The radius BO for the hemisphere (as well as of the cone) =( ½) × 4 cm = 2 cm.

So, volume of  the toy = (⅔) πr3+ (⅓) πr2h

= (⅔) × 3.14 × 23 + (⅓)× 3.14 × 22 × 2

= 25.12 cm3

Assume the right circular cylinder EFGH circumscribe the given solid.

The radius of  the base of the right circular cylinder = HP = BO = 2 cm, and the height is

EH = AO + OP = (2 + 2) cm = 4 cm

Hence, the volume required = volume of  the right circular cylinder – the volume of  the toy

= (3.14 × 22 × 4 – 25.12) cm3

= 25.12 cm3

Therefore, the required difference between the two volumes is 25.12 cm3

 

Question 66: The spherical glass vessel has a cylindrical neck of 8 cm long and 2 cm in diameter; the Diameter of  the spherical part is 8.5 cm. By measuring the amount of the water it holds, a child finds out the volume to be 345 cm3. Check if she is correct, taking the above as the inside measurements, and π = 3.14.

Answer 66:

Given,

In case of the cylinder part, Height (h) = 8 cm and Radius (R) = (2/2) cm = 1 cm

In case of the spherical part, Radius (r) = (8.5/2) = 4.25 cm

The volume of  the vessel = volume of  the cylinder + volume of  the sphere

= π × (1)2× 8 + (4/3) π (4.25)3

= 25.12 + 321.6

= 346.67 cm3

 

Question 67:  A cone of the height of 24 cm and a radius of  the base of 6 cm is made up of modelling clay. A child reshapes it in the form for the Sphere. Find out the radius of  the Sphere.

Answer 67:

The volume of  the cone = (⅓) × π × 6 × 6 × 24 cm3

When r is the radius of  the Sphere, its volume is (4/3) πr3.

As the volume of clay in the form for the cone and the Sphere remains the same, 

We have

(4/3) πr3 = (⅓) × π × 6 × 6 × 24 cm3

r3 = 3 × 3 × 24 = 33 × 23

Thus, r = 3 × 2 = 6

Hence, the radius of the Sphere is 6 cm.

 

Question 68. A cone for the radius of 8 cm and height of 12 cm is divided into two parts by a plane through the midpoint for the axis parallel to its base. Find out the ratio of the volumes of the two parts.

Answer 68:

As per the question,

height of  the cone = OM = 12 cm

The cone is divided from the mid-point.

So, let the mid-point for the cone = P

OP = PM = 6 cm

From △OPD as well as △OMN

∠POD = ∠POD [Common]

∠OPD = ∠OMN [Both 90°]

So, using the Angle-Angle similarity criterion

We have,

△OPD ~ △OMN

Also,

Similar triangles have corresponding sides in same ratio,

Hence, we have,

PD/MN = OP/OM

PD/8 = 6/12

PD = 4cm

[MN = 8 cm = radius of base of cone]

For the First part that is the cone

Base for the Radius, r = PD = 4 cm

height of  the h = OP = 6 cm

We get,

volume of  the cone for radius r and height h, V = 1/3 πr2h

volume of  the first part = 1/3 π(4)26 = 32π

For the second part, that is Frustum

Bottom for the radius, r1 = MN = 8 cm

Top for the Radius, r2 = PD = 4 cm

For Height, h = PM = 6 cm

We get,

volume of  the frustum of a cone = 1/3 πh(r12 + r22 + r1r2) , here, h = height, r1 and r2 are radii, (r1 > r2)

volume of  the second part = 1/3 π(6)[82 + 42 + 8(4)]

= 2π(112) = 224π

Hence, we get the ratio,

volume of  the first part : volume of  the second part = 32π : 224π = 1 : 7

 

Question 69. The barrel for the fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in pen is used up on writing 3300 words on an average. How many words can be written in a bottle of ink containing one-fifth for the litre?

Answer 69:

Assume first calculate the volume of  the barrel of pen that is of cylindrical shape

Consider barrel,

As 1cm = 10 mm

Base for the diameter = 5 mm = 0.5 cm

Base for the radius, r = 0.25 cm

For Height, h = 7 cm

We get,

volume of  the cylinder = πr2h

volume of  the barrel = π (0.25)27

volume of  the barrel = 22/7 × 0.25 × 0.25 × 7 = 1.375 cm3

So, as per the question,

1.375 cm3 for the ink can write 3300 words

No of words that can be written by 1 cm3 of ink = 3300/1.375 = 2400 words

1/5th of a liter = 0.2 L

We get,

1 L = 1000 cm3

0.2 L = 200 cm3

So, no for the words that can be written by 200 cm3 = 2400(200) = 480000 words

Hence, 1/5th of a litre of ink can write 480000 words.

 

Question 70. Water flows at the rate of 10m/minute and passes through a cylindrical pipe 5 mm in diameter. How long will it take to fill a conical vessel which has a diameter at the base is 40 cm and a depth of 24 cm?

Answer 70:

Assume the time taken by pipe to fill vessel = t minutes

As water flows 10 m in 1 minute, it will flow 10t meters in t minutes.

As per the question,

volume of  the conical vessel = volume of  the water that passes through the pipe in t minutes

Consider the conical pipe,

For the Base Diameter = 40 cm

For the Base radius, r = 20 cm

For Height, h = 24 cm

We know that the volume of  the cone = 1/3πr2h

volume of  the conical vessel = 1/3π(20)2(24) = 3200 π cm3

Consider cylindrical pipe,

For the Base diameter = 5 mm = 0.5 cm

For the Base radius, r = 0.25 cm

Water covers 10t m distance in pipe,

So, we get,

For Height, h = 10t m = 1000t cm

We get,

volume of  the cylinder = πr2h

volume of  the water passed in pipe = π(0.25)2(1000t) = 62.5tπ cm3

Hence, we find,

62.5tπ = 3200

62.5t = 3200

t = 51.2 minutes

We get,

0.2 minutes = 0.2(60) seconds = 12 seconds

Hence, t = 51 minutes 12 seconds

 

Question 71. A heap of rice is in the form for the cone of diameter of 9 m and height of 3.5 m. Find out the volume of  the rice. How much canvas cloth is required to cover the heap only?

Answer 71:

As per the question,

Consider conical heap,

For the Base Diameter = 9 cm

Therefore, for the base radius, r = 4.5 cm

For Height, h = 3.5 cm

We get,

Slant height,

L = √(r2+h2)

L = √(4.52+3.52)

L = √(20.25+12.25)

L = √32.5

L = 5.7cm

 The equation for the volume of cone = 1/3πr2h

We get,

volume of  the rice = volume of  the conical heap

volume of  the rice = 1/3π(4.5)2(3.5) =74.25cm3 

We also get,

Canvas requireds to just cover heap = Curved surface area of the conical heap

 curved surface area of the cone = πrl

Thus, the canvas required = π(4.5)(5.7) = 80.61 cm2 [appx]

 

Question 72. Water is flowing at the rate of 15 km/h passing through a pipe of the diameter of 14 cm into a cuboidal pond which is 50 m long and 44 m wide. At what time will the level for the water in the pond rise by 21 cm?

Answer 72:

Assume time taken by pipe to fill pond = t hours

Water flows 15 km in 1 hour, so it will flow 15t metres in t hours.

We get

volume of  the cuboidal pond up to the height of 21 cm = volume of  the water that passes through the pipe in “t” hours

Considering cuboidal pond,

For the Length, l = 50 m

For the Breadth, b = 44 m

For the Height, h = 21 cm = 0.21 m

We get,

volume of  the tank = lbh

volume of  the water = 50(44)(0.21) = 462 m3

Considering cylindrical pipe

For the Base diameter = 14 cm

For the Base radius, r = 7 cm = 0.07 m

For Height, h = 15t km = 15000t m

We get,

volume of  the cylinder = πr2h

volume of  the water passed in pipe = π(0.07)2(15000t)

=22/7 × 0.07 × 0.07 × 15000t

= 231t cm3

Thus,, we have

231t = 462

t = 2 hours

The time required to fill the tank up to a height of 25 cm is 2 hours.

 

Question 73. A solid iron cuboidal block for the dimensions 4.4 m × 2.6 m × 1m is recast into a hollow cylindrical pipe having an internal radius of 30 cm and thickness of 5 cm. Find out the length of the pipe.

Answer 73:

Consider the cuboidal block

Length, l = 4 m

Breadth, b = 2.6m  

For the Height, h = 1 m

We get

volume of  the tank = lbh

volume of  the cuboid = 4.4(2.6)(1) = 11.44 m3

We get,

The volume remains the same if a body is recast to another body.

As per the question,

volume of  the cylindrical pipe = 11.44 m3

Consider the pipe or the hollow cylinder

Internal radius of  the r2 = 30 cm = 0.3 m

Thickness = 5 cm

External radius of  the, r1 = Internal radius + thickness = 30 + 5 = 35 cm = 0.35 m

Let the length of  the pipe = h

We get,

volume of  the hollow cylinder = πh(r12– r22)

So,

volume of  the pipe = πh((0.35)2 – (0.3)2)

Hence, the length of  the pipe is 112 m.

Question 74. The metallic right circular cone of 20 cm high as well as those vertical angles is 60° cut into the two parts at the middle for the height by a plane parallel to its base. When the frustum so obtained is drawn in a wire of diameter 1/16 cm, find out the length of the wire.

Answer 74:

The diagram is as follows

Consider AEG

EG/AG =tan30°

EG = 10/√3cm = 10√3/3

In △ABD,

BD/AD = tan30°

BD = 20/√3cm =20√3/3 cm

Radius (r1) for the upper end of frustum = (10√3)/3 cm

Radius (r2) for the lower end of container = (20√3)/3 cm

Height (r3) for the container = 10 cm

Now,

volume of  the frustum = (⅓)×π×h(r12+r22+r1r2)

= (⅓)×π×10((10√3/3)2+(20√3/3)2 +(10√3/3)*(20√3/3))

 On Solving, we get,

volume of  the frustum = 22000/9 cm3

The radius (r) for the wire = (1/16)×(½) = 1/32 cm

Then,

Let the length of  the wire be “l”.

volume of  the wire = area of the cross-section x Length

= (πr2)xl

= π(1/32)2x l

Then, volume of  the frustum = volume of  the wire

22000/9 = (22/7)x(1/32)2x l

On Solving, we get

l = 7964.44 m

 

Question 75. A right triangle whose sides of 3 cm, as well as 4 cm other than hypotenuse, is made to revolve about its hypotenuse. Find out the volume and surface area of the double cone so formed. (Choose the value of π as found appropriate)

Answer 75:

Draw the diagram as follows:

consider ABA

Where,

AS = 3 cm, AC = 4 cm

Thus, Hypotenuse BC = 5 cm

We get 2 cones on the same base AA’ here the radius = DA or DA.’

Then, AD/CA = AB/CB

On substituting the value of CA, AB and CB we get,

AD = 2/5 cm

We get,

DB/AB = AB/CB

Thus, DB = 9/5 cm

Since, CD = BC-DB,

CD = 16/5 cm

volume of  the double cone will be

=[1/3π * (12/5)29/5 + 1/3π * (12/5)216/5]cm3 

On Solving this, we get,

V = 30.14 cm3

The surface area of the double cone will be

= (π*(12/5)*3)(π*(12/5)*4)cm2

=π*(12/5)*(3+4)cm2

 = 52.75 cm2

 

Question 76. Derive the formula of the curved surface area as well as total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols given.

Answer 76:

Consider the diagram,

Assume ABC be a cone. From the cone the frustum DECB is intersected by a plane parallel to its base. Where, r1 and r2 are the radii for the frustum ends for the cone and h be the frustum height.

considering the ΔABG and ΔADF,

Where, DF||BG

Therefore, ΔABG ~ ΔADF

DF/BG = AF/AG =AD/AB

r2/r1 = (h1-h)/h1 = (l1-l)/l1

r2/r1 = 1 – h/h1 = 1 – l/l1

1 – l/l1 = r2/r1

l/l1 = 1 – r2/r1 = (r1 – r2)/r1

On rearranging we find,

l1=  r1l/(r1 – r2)

CSA of frustum DECB =CSA of cone ABC -CSA cone ADE

= πr1l1– πr2(l1-l)

= πr1(r1l/(r1 – r2)) – πr2[r1l/(r1 – r2) – 1]

= πr12l/(r1 – r2) – πr2*(r1l – r1l +r2l)/(r1 – r2)

= πr12l/(r1 – r2) – πr22l/(r1 – r2)

= π[(r12– r22 )/(r1 – r2)]

CSA of frustum = π[(r1 + r2)l

 The total surface area of the frustum will be equal to the total CSA for the frustum + the area of the upper circular end + the area of the lower circular end

= π(r1+r2)l+πr22+πr22

Hence, Surface area of the frustum = π[(r1+r2)l+r12+r22]

 

Question 77. Derive the formula of the volume of the frustum for the cone.

Answer 77:

Considering the diagram 

Approach of the question in the same way as the previous one and prove that

ΔABG ~ ΔADF

Also,

DF/BG = AF/AG = AD/AB

r2/r1 = (h1-h)/h1 = (l1-l)l1

Rearranging them in the terms of h and h1

r2/r1 = 1-(h/h1) = 1 – (l/l1)

1 – (h/h1) = (r2/r1)

h/h1 = 1 – (r2/r1) = (r1-r2)/r1

h1/h = r1/(r1-r2)

h1 = r1h/(r1-r2)

The total volume of frustum for the cone will be = volume of  the cone ABC – volume of  the cone ADE

= (⅓)πr12h1 -(⅓)πr22(h1 – h)

= (π/3)[r12h1-r22(h1 – h)]

= (π/3)[r12 *hr1(r1-r2) -r22 (hr1/(r1-r2) – h)]

= (π/3)[hr13(r1-r2) -r22 (hr1– hr1 + hr2/(r1-r2))]

= (π/3)[{(hr13(r1-r2)} – {(hr23(r1-r2)}]

= (π/3)h[(r13-r23)/(r13-r23)]

= (π/3)h[(r1-r2)(r12+r22+r1r2)/(r1-r2)]

On solving, we get,

Thus,  Volume of frustum for the cone = (⅓)πh(r12+r22+r1r2)

Question 78: An open metal bucket is at the shape of the frustum for the cone, mounted on a hollow cylindrical base made for the same metallic sheet as shown in the figure. The diameters for the two circular ends of the bucket are 45 cm and 25 cm, and the total vertical height of  the bucket is 40 cm, as well as that of the cylindrical base is 6 cm. Find out the area of the metallic sheet used to make the bucket. Here, we do not take into account the handle for the bucket. Also, find out the volume of water the bucket can hold. (Take π = 22/7)

Answer 78:

The total height of  the bucket = 40 cm, which includes the height of the base.

Hence, the height of  the frustum of the cone = (40 – 6) cm = 34 cm.

Therefore, the slant height of  the frustum,

here r1 = 22.5 cm, r2 = 12.5 cm and h = 34 cm.

=35.44 cm

The area of the metallic sheet used = curved surface area of the frustum of cone + area of the circular base + curved surface area of the cylinder

= [π × 35.44 (22.5 + 12.5) + π × (12.5)2 + 2π × 12.5 × 6] cm2

= 22/7 (1240.4 + 156.25 + 150) cm2

= 4860.9 cm2

Then,, the volume of  the water that the bucket can be hold also, called  capacity

of the bucket

= (πh)/3 × (r12 + r22+ r1r2)

=(22/7) × (34/3) × [(22.5)2 + (12.5)2 + (22.5)(12.5)]

=(22/7) × (34/3) × 943.75

= 33615.48 cm3

= 33.62 litres (approximately.)

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